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I'm new to digital radios and signal processing, so I apologize if this question is trivial but I haven't been able to find an answer here or by googling. Also, some terminology might be off, please feel free to refer me to correct sources or to correct my basic understanding.

Reading various sources (e.g. here), it seems to me that the I and Q components of a sample correspond to the complex representation of a portion of a sine wave described by $I \cdot \cos(2 \pi f t) + Q \cdot \sin(2 \pi f t)$ w.r.t. $t$, where $f$ denotes the frequency of interest. My question is, how does the receiver actually compute $I$ and $Q$ when a sample is needed?

Suppose that a sample is taken at a time $t$, I don't think that the receiver could just multiply the instantaneous strength $V$ (voltage?) of the incoming signal by $\cos(2\pi ft)$ and by $\sin(2 \pi f t)$ to recover $I$ and $Q$ (as the diagram in section "Receiver Side" of the linked article appears to suggests) since this would carry no more information than reporting $V$ itself.

Moreover, in principle, the incoming voltage from the antenna on the receiver side could be any continuous (and differentiable?) function $V(t)$... so how are $I$ and $Q$ recovered? Are they actually the values that minimize some error function between the incoming voltage and the function described by $I \cdot \sin(f) + Q \cdot \cos(f)$ over a length of time corresponding to some sampling interval $[t, t']$? E.g. something along the lines of: $$ I,Q = \arg\min_{I,Q \in \mathbb{R}}\int_{\tau=t}^{t'} \big( I \cdot \cos(2 \pi f \tau) + Q \cdot \sin(2 \pi f \tau) - V(\tau) \big)^2 \;\mbox{d}\tau \;\mbox{ ?} $$

Thank you!

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    $\begingroup$ Hello Steven, and welcome to ham.stackexchange.com! $\endgroup$
    – rclocher3
    Dec 17 '20 at 0:01
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Suppose that a sample is taken at a time $t$, I don't think that the receiver could just multiply the instantaneous strength $V$ (voltage?) of the incoming signal by $\cos(2\pi ft)$ and by $\sin(2 \pi f t)$ to recover $I$ and $Q$ (as the diagram in section "Receiver Side" of the linked article appears to suggests) since this would carry no more information than reporting $V$ itself.

It can, and it does precisely this. But you are right that it carries no more information.

In practice it carries less, and that's the point. Say we want to make a WiFi radio operating in the 5 GHz band. This would require a sample rate of at least 10 GHz. That would be an expensive ADC, as would the computing power to process such a high sample rate.

But the bandwidth of a WiFi signal is only some 10s of MHz. The point of the mixer is to convert the signal at high frequency (somewhere in the 5 GHz band) down to a lower frequency which can be represented at a lower sample rate and thus more easily digitized and processed.

So, the output of the mixer is low-pass filtered before being digitized by the ADC.

Moreover, in principle, the incoming voltage from the antenna on the receiver side could be any continuous (and differentiable?) function $V(t)$... so how are $I$ and $Q$ recovered? Are they actually the values that minimize some error function [...]

No, it's nothing so complex. Remember the mixer is an analog component, so there's no need for any "sampling interval", and an arbitrary continuous function is no problem. The ideal mixer performs simply:

$$ I = V(t) \cdot \cos(2\pi f) \\ Q = V(t) \cdot \sin(2\pi f) $$

If I and Q are interpreted as the real and imaginary parts of a complex number respectively, it is simpler (by Euler's formula) to think of the mixer as performing:

$$ V(t) \cdot e^{i 2 \pi f} $$

This is useful because multiplying by $e^{i 2 \pi f}$ shifts all frequencies by $f$, which you can see for example in rule 103 of Wikipedia's list of Fourier transforms.

These analog signals are then low-pass filtered and digitized by the ADC.

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  • $\begingroup$ Thank you very much. So in principle one could just shift down the signal so that the frequency of interest is near 0Hz, then report the instantaneous amplitude at about twice the bandwidth per second. Is this is correct, why is this not done? Is it just that one easily gets $I$ and $Q$ as a byproduct of this downshifting process using the quadrature mixer, and so it would actually be extra work to re-convert $I$ and $Q$ to a single sample? Wouldn't this extra step halve the bandwidth needed to report samples between the receiver and software? $\endgroup$
    – Steven
    Dec 17 '20 at 10:05
  • $\begingroup$ @Steven Does ham.stackexchange.com/q/10454/218 answer your question? If not, please do ask another. $\endgroup$ Dec 17 '20 at 13:28
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the I and Q components of a sample correspond to the complex representation of a portion of a sine wave described by $I \cdot \cos(2 \pi f t) + Q \cdot \sin(2 \pi f t)$ w.r.t. $t$, where $f$ denotes the frequency of interest

This is correct (if we suppose the incoming signal is a sine wave, i.e. an unmodulated carrier).

I don't think that the receiver could just multiply the instantaneous strength $V$ (voltage?) of the incoming signal by $\cos(2\pi ft)$ and by $\sin(2 \pi f t)$ to recover $I$ and $Q$ … since this would carry no more information than reporting $V$ itself.

Actually, this is useful. The key facts are:

  • This multiplication can done in the analog domain, using a quadrature mixer, to produce a new pair of “downconverted” signals without sampling them yet. This is how SDRs avoid needing gigahertz-rate analog-to-digital conversion.
  • A signal of actually interesting content (modulation) is not just a pure sine wave, but has other frequency components.

These I and Q signals have had all their frequency components shifted down in frequency by $f$ — this is known as “baseband”. The signals are then low-pass filtered (which removes all frequencies outside of the range $f ± \text{filter frequency}$ in the original signal) and sampled by an ADC to produce the digital baseband signal.

Note that this means that an incoming signal at frequency $f$ has frequency zero in the baseband representation. If the signal is a sine wave with a small difference from $f$ (e.g. perhaps it is frequency-modulated around $f$) then the baseband form has a small difference from zero. If it has more frequency components, all of those are still present in the baseband signal, just translated.

You're correct to think that an IQ form of the original RF signal contains no more information than the original instantaneous voltage. The point of IQ is to allow us to throw out something we don't need — the extremely high carrier frequency $f$ — without discarding the information we care about in the signal (provided it is limited to a small band around $f$), so as to be able to receive, digitize, and demodulate it with simple general-purpose hardware.

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  • $\begingroup$ I don't think we need to assume the signal is just a sine wave for the first point to be true. Notice this is effectively multiplication by $e^{i2\pi ft}$, that is, a shift in frequency. It's what the SDR does to convert the signal to baseband, and flipping the sign of $f$ performs the reverse operation. Works for any function, not just sinusoids. $\endgroup$ Dec 17 '20 at 3:14
  • $\begingroup$ @PhilFrost-W8II Steven wrote "the I and Q components of a sample correspond to the complex representation of a portion of a sine wave". Assuming "correspond to" means something like "proportional to", this is true only if the signal being sampled is a sine wave. $\endgroup$
    – Kevin Reid AG6YO
    Dec 17 '20 at 5:50
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In most typical SDR receivers, I and Q are not determined from the instantaneous RF voltage input, but from a reduced bandwidth slice of the RF spectrum. The slice is taken by quadrature heterodyning/mixing (with a quadrature local oscillator (LO) near the frequency slice of interest), thus producing two signals. This pair of mixer results is typically low pass filtered, then sampled by 2 ADCs, usually at a much lower rate than the LO frequency, to produce sampled IQ data suitable for software processing. The low pass filtering plus sampling thus sort-of averages the RF within a certain band or slice, but with two different or offset time comb windows (the 2 quadrature mixer LO inputs), thus producing I and Q magnitude and phase information about all the various signals within the band-limited spectrum slice.

A direct sampling SDR receiver also does the above, but inverts the order of the mixing and the ADC sampling to sample first then quadrature mix (then filter and decimate digitally, perhaps in an FPGA). The mixing and filtering can also be done in multiple stages, some in hardware/gateware, some in software, using multiple quadrature LOs, multiple filter stages, and digital complex multiplication.

If you want to use that integral, it needs to integrate over a window function that is a composite of the impulse response of the low pass filter(s) and the capture window(s) of the ADC(s). For each sample. For each of I and Q.

No instantaneous voltages are measured (because capacitance in the real world requires finite time to charge up or down to any measurable level).

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    $\begingroup$ What's an "offset time comb window"? $\endgroup$ Dec 16 '20 at 22:15
  • $\begingroup$ The 2 LO inputs to the mixer can either be (Tayloe) square waves, which are an obvious comb, or sinusoids, which has comb-like zero crossings. The relative offset of the 2 quadrature components select which parts of the RF mixer input they (mostly) ignore at 2 different sets of times. $\endgroup$
    – hotpaw2
    Dec 17 '20 at 17:41
  • $\begingroup$ I'm sorry but I don't understand how a sinusoid can be "comb-like" in any way, when the quadrature components would have any offset other than 90 degrees ("quadrature"), or what 2 different sets of times we are talking about. $\endgroup$ Dec 17 '20 at 19:55
  • $\begingroup$ At all zero crossings, a sinusoidal input to a mixer causes zero output. And very diminished output near those zero crossings as well. What's non comb-like about a continuous periodic sequence of zero crossings. And the more common Tayloe mixers in most QRP SDRs use square waves, not sinusoids. $\endgroup$
    – hotpaw2
    Dec 17 '20 at 22:48
  • $\begingroup$ That's certainly a novel way to use the word "comb", but OK. $\endgroup$ Dec 17 '20 at 23:04

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